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julia-pushkina [17]
2 years ago
15

Are fossils likely to form in areas that experience a lot of erosion on daily basis?

Chemistry
1 answer:
frosja888 [35]2 years ago
6 0
No, There are not likely to form in areas that experience a lot of erosion on a daily basis.
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By hot springs, fumaroles and geysers.
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Mass number is equal to
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State whether the following statements are true or false (with justification). (a) 1 mol of N2 has more molecules than 1 mol of
MAVERICK [17]

Answer:

A. False.

Every substance contains the same number of molecules i.e 6.02x10^23 molecules

B. False.

Mass conc. = number mole x molar Mass

Mass conc. of 1mole of N2 = 1 x 28 = 28g

Mass conc. of 1mol of Ar = 1 x 40 = 40g

The mass of 1mole of Ar is greater than the mass of 1mole of N2

C. False.

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Explanation:

7 0
3 years ago
The ionic equation for MgCl2 and KOH.
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The reaction of Mg Cl2 and KOH can be described as a double substitution type of reaction. This means the cations of the reactants are exchanged in places when the products are formed. In this case, the balanced reaction is expressed
MgCl2 (s) + 2KOH (aq) = Mhg (OH)2 (aq) + 2KCl (s)
5 0
3 years ago
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What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

So, the pH of the buffer is 6.1236.

5 0
3 years ago
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